My friends and I likes to play Trivial Pursuit without using the board.

We play it like this:

  • Throw a die to determine what color you get to answer.
  • Ask a question, if the answer is correct you get a point.
  • If enough points are awarded you win

We would like to modify the game as to include the colors. There are 6 colors. The game could then be won by completing all colors or answering enough questions.

We would like the the effort to complete it by numbers to be similar to that of completing it by colors. So the required number of correct answers should be the same as where it is likely that all the colors has been collected.

What is the number of correct answers one needs to acquire to make it probable, P>=0.5, that all colors are collected?

We dabbled in a few sums before realizing this was over our heads.

  • $\begingroup$ Wouldn't this depend on the probability of answering a question correctly? $\endgroup$ – user5137 Oct 28 '11 at 17:40
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    $\begingroup$ Are you assuming that one is equally good at answering questions of any color? The more lopsided one's abilities are, the longer will it take to collect all colors, relative to just raking up a given number of successes. $\endgroup$ – Henning Makholm Oct 28 '11 at 17:41
  • $\begingroup$ @Henning Yes, exactly, equal probability for answering any color correctly. $\endgroup$ – Archimedes Oct 28 '11 at 17:43

This is the coupon collector's problem. For six, on average you will need $6/6+6/5+6/4+6/3+6/2+6/1=14.7$ correct answers, but the variability is high. This is the expectation, not the number to have 50% chance of success.

  • $\begingroup$ The wikipedia article makes it apparent that this is indeed the correct solution. Is there any intuitive way to understand this? Our own intuition was guessing in the range 25-30 correct answers. $\endgroup$ – Archimedes Oct 28 '11 at 17:53
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    $\begingroup$ Yes, each fraction is the average time to get the next different coupon. The first coupon is guaranteed to be new. After you have 1, 5 of 6 coupons will be new, so on average it takes 6/5 tries to get a new one. After you have 5, on average it takes 6 tries to get the last one. $\endgroup$ – Ross Millikan Oct 28 '11 at 17:55
  • $\begingroup$ Ah, thinking about it in terms of time/tries does help a lot. Thank you! When I get 15 rep you'll get the upvote=) $\endgroup$ – Archimedes Oct 28 '11 at 17:57
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    $\begingroup$ @Archimedes: Here are two previous posts that deal with the same or very similar questions: math.stackexchange.com/questions/28905/…, math.stackexchange.com/questions/57547/…. $\endgroup$ – joriki Oct 28 '11 at 18:00
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    $\begingroup$ @Archimedes Heh, +1 to your question so you will have 15 rep. It's also an interesting question. $\endgroup$ – Graphth Oct 28 '11 at 18:02

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